点与线段之间的最短距离

Question

我需要一个基本函数来找到点和线段之间的最短距离。 随意用您想要的任何语言编写解决方案； 我可以将其翻译成我正在使用的（Javascript）。

编辑：我的线段由两个端点定义。 因此，我的线段 AB 是由两点 A (x1,y1) 和 B (x2,y2) 定义的。 我试图找到该线段和点 C (x3,y3) 之间的距离。 我很遗憾地承认，我的几何技能很生疏，所以我看到的例子很令人困惑。

Answer 1

WPF版本：

public class LineSegment
{
    private readonly Vector _offset;
    private readonly Vector _vector;

    public LineSegment(Point start, Point end)
    {
        _offset = (Vector)start;
        _vector = (Vector)(end - _offset);
    }

    public double DistanceTo(Point pt)
    {
        var v = (Vector)pt - _offset;

        // first, find a projection point on the segment in parametric form (0..1)
        var p = (v * _vector) / _vector.LengthSquared;

        // and limit it so it lays inside the segment
        p = Math.Min(Math.Max(p, 0), 1);

        // now, find the distance from that point to our point
        return (_vector * p - v).Length;
    }
}

Answer 2

无法抗拒用 python 进行编码:)

from math import sqrt, fabs
def pdis(a, b, c):
    t = b[0]-a[0], b[1]-a[1]           # Vector ab
    dd = sqrt(t[0]**2+t[1]**2)         # Length of ab
    t = t[0]/dd, t[1]/dd               # unit vector of ab
    n = -t[1], t[0]                    # normal unit vector to ab
    ac = c[0]-a[0], c[1]-a[1]          # vector ac
    return fabs(ac[0]*n[0]+ac[1]*n[1]) # Projection of ac to n (the minimum distance)

print pdis((1,1), (2,2), (2,0))        # Example (answer is 1.414)

Fortran 也是如此:)

real function pdis(a, b, c)
    real, dimension(0:1), intent(in) :: a, b, c
    real, dimension(0:1) :: t, n, ac
    real :: dd
    t = b - a                          ! Vector ab
    dd = sqrt(t(0)**2+t(1)**2)         ! Length of ab
    t = t/dd                           ! unit vector of ab
    n = (/-t(1), t(0)/)                ! normal unit vector to ab
    ac = c - a                         ! vector ac
    pdis = abs(ac(0)*n(0)+ac(1)*n(1))  ! Projection of ac to n (the minimum distance)
end function pdis


program test
    print *, pdis((/1.0,1.0/), (/2.0,2.0/), (/2.0,0.0/))   ! Example (answer is 1.414)
end program test

Answer 3

这是 Grumdrig 解决方案的更完整拼写。 此版本还返回最近的点本身。

#include "stdio.h"
#include "math.h"

class Vec2
{
public:
    float _x;
    float _y;

    Vec2()
    {
        _x = 0;
        _y = 0;
    }

    Vec2( const float x, const float y )
    {
        _x = x;
        _y = y;
    }

    Vec2 operator+( const Vec2 &v ) const
    {
        return Vec2( this->_x + v._x, this->_y + v._y );
    }

    Vec2 operator-( const Vec2 &v ) const
    {
        return Vec2( this->_x - v._x, this->_y - v._y );
    }

    Vec2 operator*( const float f ) const
    {
        return Vec2( this->_x * f, this->_y * f );
    }

    float DistanceToSquared( const Vec2 p ) const
    {
        const float dX = p._x - this->_x;
        const float dY = p._y - this->_y;

        return dX * dX + dY * dY;
    }

    float DistanceTo( const Vec2 p ) const
    {
        return sqrt( this->DistanceToSquared( p ) );
    }

    float DotProduct( const Vec2 p ) const
    {
        return this->_x * p._x + this->_y * p._y;
    }
};

// return minimum distance between line segment vw and point p, and the closest point on the line segment, q
float DistanceFromLineSegmentToPoint( const Vec2 v, const Vec2 w, const Vec2 p, Vec2 * const q )
{
    const float distSq = v.DistanceToSquared( w ); // i.e. |w-v|^2 ... avoid a sqrt
    if ( distSq == 0.0 )
    {
        // v == w case
        (*q) = v;

        return v.DistanceTo( p );
    }

    // consider the line extending the segment, parameterized as v + t (w - v)
    // we find projection of point p onto the line
    // it falls where t = [(p-v) . (w-v)] / |w-v|^2

    const float t = ( p - v ).DotProduct( w - v ) / distSq;
    if ( t < 0.0 )
    {
        // beyond the v end of the segment
        (*q) = v;

        return v.DistanceTo( p );
    }
    else if ( t > 1.0 )
    {
        // beyond the w end of the segment
        (*q) = w;

        return w.DistanceTo( p );
    }

    // projection falls on the segment
    const Vec2 projection = v + ( ( w - v ) * t );

    (*q) = projection;

    return p.DistanceTo( projection );
}

float DistanceFromLineSegmentToPoint( float segmentX1, float segmentY1, float segmentX2, float segmentY2, float pX, float pY, float *qX, float *qY )
{
    Vec2 q;

    float distance = DistanceFromLineSegmentToPoint( Vec2( segmentX1, segmentY1 ), Vec2( segmentX2, segmentY2 ), Vec2( pX, pY ), &q );

    (*qX) = q._x;
    (*qY) = q._y;

    return distance;
}

void TestDistanceFromLineSegmentToPoint( float segmentX1, float segmentY1, float segmentX2, float segmentY2, float pX, float pY )
{
    float qX;
    float qY;
    float d = DistanceFromLineSegmentToPoint( segmentX1, segmentY1, segmentX2, segmentY2, pX, pY, &qX, &qY );
    printf( "line segment = ( ( %f, %f ), ( %f, %f ) ), p = ( %f, %f ), distance = %f, q = ( %f, %f )\n",
            segmentX1, segmentY1, segmentX2, segmentY2, pX, pY, d, qX, qY );
}

void TestDistanceFromLineSegmentToPoint()
{
    TestDistanceFromLineSegmentToPoint( 0, 0, 1, 1, 1, 0 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, 5, 4 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, 30, 15 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, -30, 15 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 10, 0, 5, 1 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 0, 10, 1, 5 );
}

Answer 4

我假设您想找到点和线段之间的最短距离； 为此，您需要找到与穿过您的点的线段（lineB）垂直的线（lineA），确定该线（lineA）与穿过您的线段（lineB）的线之间的交点; 如果该点位于线段的两点之间，则该距离是您的点与刚刚找到的点（即 lineA 和 lineB 的交点）之间的距离； 如果该点不在线段两点之间，则需要获取该点与线段两端较近的点之间的距离； 这可以通过取该点和线段两点之间的平方距离（以避免平方根）来轻松完成； 以更接近的为准，取其平方根。

Answer 5

考虑对上面格鲁姆德里格的答案的修改。 很多时候，您会发现浮点不精确会导致问题。 我在下面的版本中使用双精度数，但您可以轻松更改为浮点数。 重要的部分是它使用 epsilon 来处理“slop”。 此外，您很多时候想知道交叉点发生在哪里，或者是否发生过。 如果返回的t是< 0.0或> 1.0，没有发生碰撞。 然而，即使没有发生碰撞，很多时候你也会想知道线段上距离 P 最近的点在哪里，因此我使用 qx 和 qy 来返回该位置。

double PointSegmentDistanceSquared( double px, double py,
                                    double p1x, double p1y,
                                    double p2x, double p2y,
                                    double& t,
                                    double& qx, double& qy)
{
    static const double kMinSegmentLenSquared = 0.00000001;  // adjust to suit.  If you use float, you'll probably want something like 0.000001f
    static const double kEpsilon = 1.0E-14;  // adjust to suit.  If you use floats, you'll probably want something like 1E-7f
    double dx = p2x - p1x;
    double dy = p2y - p1y;
    double dp1x = px - p1x;
    double dp1y = py - p1y;
    const double segLenSquared = (dx * dx) + (dy * dy);
    if (segLenSquared >= -kMinSegmentLenSquared && segLenSquared <= kMinSegmentLenSquared)
    {
        // segment is a point.
        qx = p1x;
        qy = p1y;
        t = 0.0;
        return ((dp1x * dp1x) + (dp1y * dp1y));
    }
    else
    {
        // Project a line from p to the segment [p1,p2].  By considering the line
        // extending the segment, parameterized as p1 + (t * (p2 - p1)),
        // we find projection of point p onto the line. 
        // It falls where t = [(p - p1) . (p2 - p1)] / |p2 - p1|^2
        t = ((dp1x * dx) + (dp1y * dy)) / segLenSquared;
        if (t < kEpsilon)
        {
            // intersects at or to the "left" of first segment vertex (p1x, p1y).  If t is approximately 0.0, then
            // intersection is at p1.  If t is less than that, then there is no intersection (i.e. p is not within
            // the 'bounds' of the segment)
            if (t > -kEpsilon)
            {
                // intersects at 1st segment vertex
                t = 0.0;
            }
            // set our 'intersection' point to p1.
            qx = p1x;
            qy = p1y;
            // Note: If you wanted the ACTUAL intersection point of where the projected lines would intersect if
            // we were doing PointLineDistanceSquared, then qx would be (p1x + (t * dx)) and qy would be (p1y + (t * dy)).
        }
        else if (t > (1.0 - kEpsilon))
        {
            // intersects at or to the "right" of second segment vertex (p2x, p2y).  If t is approximately 1.0, then
            // intersection is at p2.  If t is greater than that, then there is no intersection (i.e. p is not within
            // the 'bounds' of the segment)
            if (t < (1.0 + kEpsilon))
            {
                // intersects at 2nd segment vertex
                t = 1.0;
            }
            // set our 'intersection' point to p2.
            qx = p2x;
            qy = p2y;
            // Note: If you wanted the ACTUAL intersection point of where the projected lines would intersect if
            // we were doing PointLineDistanceSquared, then qx would be (p1x + (t * dx)) and qy would be (p1y + (t * dy)).
        }
        else
        {
            // The projection of the point to the point on the segment that is perpendicular succeeded and the point
            // is 'within' the bounds of the segment.  Set the intersection point as that projected point.
            qx = p1x + (t * dx);
            qy = p1y + (t * dy);
        }
        // return the squared distance from p to the intersection point.  Note that we return the squared distance
        // as an optimization because many times you just need to compare relative distances and the squared values
        // works fine for that.  If you want the ACTUAL distance, just take the square root of this value.
        double dpqx = px - qx;
        double dpqy = py - qy;
        return ((dpqx * dpqx) + (dpqy * dpqy));
    }
}

Answer 6

Grumdrig 的 C++/JavaScript 实现对我来说非常有用，因此我提供了我正在使用的 Python 直接端口。 完整的代码位于此处。

class Point(object):
  def __init__(self, x, y):
    self.x = float(x)
    self.y = float(y)

def square(x):
  return x * x

def distance_squared(v, w):
  return square(v.x - w.x) + square(v.y - w.y)

def distance_point_segment_squared(p, v, w):
  # Segment length squared, |w-v|^2
  d2 = distance_squared(v, w) 
  if d2 == 0: 
    # v == w, return distance to v
    return distance_squared(p, v)
  # Consider the line extending the segment, parameterized as v + t (w - v).
  # We find projection of point p onto the line.
  # It falls where t = [(p-v) . (w-v)] / |w-v|^2
  t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / d2;
  if t < 0:
    # Beyond v end of the segment
    return distance_squared(p, v)
  elif t > 1.0:
    # Beyond w end of the segment
    return distance_squared(p, w)
  else:
    # Projection falls on the segment.
    proj = Point(v.x + t * (w.x - v.x), v.y + t * (w.y - v.y))
    # print proj.x, proj.y
    return distance_squared(p, proj)

Answer 7

这里使用的是 Swift

    /* Distance from a point (p1) to line l1 l2 */
func distanceFromPoint(p: CGPoint, toLineSegment l1: CGPoint, and l2: CGPoint) -> CGFloat {
    let A = p.x - l1.x
    let B = p.y - l1.y
    let C = l2.x - l1.x
    let D = l2.y - l1.y
    
    let dot = A * C + B * D
    let len_sq = C * C + D * D
    var param : CGFloat = -1
    if len_sq != 0 {
        param = dot / len_sq
    }
    
    var xx, yy: CGFloat
    
    if param < 0 || (l1.x == l2.x && l1.y == l2.y) {
        xx = l1.x
        yy = l1.y
    } else if param > 1 {
        xx = l2.x
        yy = l2.y
    } else {
        xx = l1.x + param * C
        yy = l1.y + param * D
    }
    
    let dx = p.x - xx
    let dy = p.y - yy
    
    return sqrt(dx * dx + dy * dy)
}

Answer 8

现在我的解决方案也是......
（Javascript）

它非常快，因为我试图避免任何 Math.pow 函数。

正如你所看到的，在函数的末尾我有直线的距离。

代码来自lib http://www.draw2d.org/graphiti/ jsdoc/#!/example

/**
 * Static util function to determine is a point(px,py) on the line(x1,y1,x2,y2)
 * A simple hit test.
 * 
 * @return {boolean}
 * @static
 * @private
 * @param {Number} coronaWidth the accepted corona for the hit test
 * @param {Number} X1 x coordinate of the start point of the line
 * @param {Number} Y1 y coordinate of the start point of the line
 * @param {Number} X2 x coordinate of the end point of the line
 * @param {Number} Y2 y coordinate of the end point of the line
 * @param {Number} px x coordinate of the point to test
 * @param {Number} py y coordinate of the point to test
 **/
graphiti.shape.basic.Line.hit= function( coronaWidth, X1, Y1,  X2,  Y2, px, py)
{
  // Adjust vectors relative to X1,Y1
  // X2,Y2 becomes relative vector from X1,Y1 to end of segment
  X2 -= X1;
  Y2 -= Y1;
  // px,py becomes relative vector from X1,Y1 to test point
  px -= X1;
  py -= Y1;
  var dotprod = px * X2 + py * Y2;
  var projlenSq;
  if (dotprod <= 0.0) {
      // px,py is on the side of X1,Y1 away from X2,Y2
      // distance to segment is length of px,py vector
      // "length of its (clipped) projection" is now 0.0
      projlenSq = 0.0;
  } else {
      // switch to backwards vectors relative to X2,Y2
      // X2,Y2 are already the negative of X1,Y1=>X2,Y2
      // to get px,py to be the negative of px,py=>X2,Y2
      // the dot product of two negated vectors is the same
      // as the dot product of the two normal vectors
      px = X2 - px;
      py = Y2 - py;
      dotprod = px * X2 + py * Y2;
      if (dotprod <= 0.0) {
          // px,py is on the side of X2,Y2 away from X1,Y1
          // distance to segment is length of (backwards) px,py vector
          // "length of its (clipped) projection" is now 0.0
          projlenSq = 0.0;
      } else {
          // px,py is between X1,Y1 and X2,Y2
          // dotprod is the length of the px,py vector
          // projected on the X2,Y2=>X1,Y1 vector times the
          // length of the X2,Y2=>X1,Y1 vector
          projlenSq = dotprod * dotprod / (X2 * X2 + Y2 * Y2);
      }
  }
    // Distance to line is now the length of the relative point
    // vector minus the length of its projection onto the line
    // (which is zero if the projection falls outside the range
    //  of the line segment).
    var lenSq = px * px + py * py - projlenSq;
    if (lenSq < 0) {
        lenSq = 0;
    }
    return Math.sqrt(lenSq)<coronaWidth;
};

Answer 9

C#

改编自@Grumdrig

public static double MinimumDistanceToLineSegment(this Point p,
    Line line)
{
    var v = line.StartPoint;
    var w = line.EndPoint;

    double lengthSquared = DistanceSquared(v, w);

    if (lengthSquared == 0.0)
        return Distance(p, v);

    double t = Math.Max(0, Math.Min(1, DotProduct(p - v, w - v) / lengthSquared));
    var projection = v + t * (w - v);

    return Distance(p, projection);
}

public static double Distance(Point a, Point b)
{
    return Math.Sqrt(DistanceSquared(a, b));
}

public static double DistanceSquared(Point a, Point b)
{
    var d = a - b;
    return DotProduct(d, d);
}

public static double DotProduct(Point a, Point b)
{
    return (a.X * b.X) + (a.Y * b.Y);
}

Answer 10

Matlab 代码，如果调用不带参数的函数，则具有内置的“自测试”：

function r = distPointToLineSegment( xy0, xy1, xyP )
% r = distPointToLineSegment( xy0, xy1, xyP )

if( nargin < 3 )
    selfTest();
    r=0;
else
    vx = xy0(1)-xyP(1);
    vy = xy0(2)-xyP(2);
    ux = xy1(1)-xy0(1);
    uy = xy1(2)-xy0(2);
    lenSqr= (ux*ux+uy*uy);
    detP= -vx*ux + -vy*uy;

    if( detP < 0 )
        r = norm(xy0-xyP,2);
    elseif( detP > lenSqr )
        r = norm(xy1-xyP,2);
    else
        r = abs(ux*vy-uy*vx)/sqrt(lenSqr);
    end
end


    function selfTest()
        %#ok<*NASGU>
        disp(['invalid args, distPointToLineSegment running (recursive)  self-test...']);

        ptA = [1;1]; ptB = [-1;-1];
        ptC = [1/2;1/2];  % on the line
        ptD = [-2;-1.5];  % too far from line segment
        ptE = [1/2;0];    % should be same as perpendicular distance to line
        ptF = [1.5;1.5];      % along the A-B but outside of the segment

        distCtoAB = distPointToLineSegment(ptA,ptB,ptC)
        distDtoAB = distPointToLineSegment(ptA,ptB,ptD)
        distEtoAB = distPointToLineSegment(ptA,ptB,ptE)
        distFtoAB = distPointToLineSegment(ptA,ptB,ptF)
        figure(1); clf;
        circle = @(x, y, r, c) rectangle('Position', [x-r, y-r, 2*r, 2*r], ...
            'Curvature', [1 1], 'EdgeColor', c);
        plot([ptA(1) ptB(1)],[ptA(2) ptB(2)],'r-x'); hold on;
        plot(ptC(1),ptC(2),'b+'); circle(ptC(1),ptC(2), 0.5e-1, 'b');
        plot(ptD(1),ptD(2),'g+'); circle(ptD(1),ptD(2), distDtoAB, 'g');
        plot(ptE(1),ptE(2),'k+'); circle(ptE(1),ptE(2), distEtoAB, 'k');
        plot(ptF(1),ptF(2),'m+'); circle(ptF(1),ptF(2), distFtoAB, 'm');
        hold off;
        axis([-3 3 -3 3]); axis equal;
    end

end

Answer 11

用t-sql编码，

点是（@px，@py），线段从（@ax，@ay）到（@bx，@by）

create function fn_sqr (@NumberToSquare decimal(18,10)) 
returns decimal(18,10)
as 
begin
    declare @Result decimal(18,10)
    set @Result = @NumberToSquare * @NumberToSquare
    return @Result
end
go

create function fn_Distance(@ax decimal (18,10) , @ay decimal (18,10), @bx decimal(18,10),  @by decimal(18,10)) 
returns decimal(18,10)
as
begin
    declare @Result decimal(18,10)
    set @Result = (select dbo.fn_sqr(@ax - @bx) + dbo.fn_sqr(@ay - @by) )
    return @Result
end
go

create function fn_DistanceToSegmentSquared(@px decimal(18,10), @py decimal(18,10), @ax decimal(18,10), @ay decimal(18,10), @bx decimal(18,10), @by decimal(18,10)) 
returns decimal(18,10)
as 
begin
    declare @l2 decimal(18,10)
    set @l2 = (select dbo.fn_Distance(@ax, @ay, @bx, @by))
    if @l2 = 0
        return dbo.fn_Distance(@px, @py, @ax, @ay)
    declare @t decimal(18,10)
    set @t = ((@px - @ax) * (@bx - @ax) + (@py - @ay) * (@by - @ay)) / @l2
    if (@t < 0) 
        return dbo.fn_Distance(@px, @py, @ax, @ay);
    if (@t > 1) 
        return dbo.fn_Distance(@px, @py, @bx, @by);
    return dbo.fn_Distance(@px, @py,  @ax + @t * (@bx - @ax),  @ay + @t * (@by - @ay))
end
go

create function fn_DistanceToSegment(@px decimal(18,10), @py decimal(18,10), @ax decimal(18,10), @ay decimal(18,10), @bx decimal(18,10), @by decimal(18,10)) 
returns decimal(18,10)
as 
begin
    return sqrt(dbo.fn_DistanceToSegmentSquared(@px, @py , @ax , @ay , @bx , @by ))
end
go

--example execution for distance from a point at (6,1) to line segment that runs from (4,2) to (2,1)
select dbo.fn_DistanceToSegment(6, 1, 4, 2, 2, 1) 
--result = 2.2360679775

--example execution for distance from a point at (-3,-2) to line segment that runs from (0,-2) to (-2,1)
select dbo.fn_DistanceToSegment(-3, -2, 0, -2, -2, 1) 
--result = 2.4961508830

--example execution for distance from a point at (0,-2) to line segment that runs from (0,-2) to (-2,1)
select dbo.fn_DistanceToSegment(0,-2, 0, -2, -2, 1) 
--result = 0.0000000000

Answer 12

看起来 StackOverflow 上的几乎所有其他人都贡献了答案（到目前为止有 23 个答案），所以这是我对 C# 的贡献。 这主要基于 M. Katz 的答案，而 M. Katz 的答案又基于 Grumdrig 的答案。

   public struct MyVector
   {
      private readonly double _x, _y;


      // Constructor
      public MyVector(double x, double y)
      {
         _x = x;
         _y = y;
      }


      // Distance from this point to another point, squared
      private double DistanceSquared(MyVector otherPoint)
      {
         double dx = otherPoint._x - this._x;
         double dy = otherPoint._y - this._y;
         return dx * dx + dy * dy;
      }


      // Find the distance from this point to a line segment (which is not the same as from this 
      //  point to anywhere on an infinite line). Also returns the closest point.
      public double DistanceToLineSegment(MyVector lineSegmentPoint1, MyVector lineSegmentPoint2,
                                          out MyVector closestPoint)
      {
         return Math.Sqrt(DistanceToLineSegmentSquared(lineSegmentPoint1, lineSegmentPoint2, 
                          out closestPoint));
      }


      // Same as above, but avoid using Sqrt(), saves a new nanoseconds in cases where you only want 
      //  to compare several distances to find the smallest or largest, but don't need the distance
      public double DistanceToLineSegmentSquared(MyVector lineSegmentPoint1, 
                                              MyVector lineSegmentPoint2, out MyVector closestPoint)
      {
         // Compute length of line segment (squared) and handle special case of coincident points
         double segmentLengthSquared = lineSegmentPoint1.DistanceSquared(lineSegmentPoint2);
         if (segmentLengthSquared < 1E-7f)  // Arbitrary "close enough for government work" value
         {
            closestPoint = lineSegmentPoint1;
            return this.DistanceSquared(closestPoint);
         }

         // Use the magic formula to compute the "projection" of this point on the infinite line
         MyVector lineSegment = lineSegmentPoint2 - lineSegmentPoint1;
         double t = (this - lineSegmentPoint1).DotProduct(lineSegment) / segmentLengthSquared;

         // Handle the two cases where the projection is not on the line segment, and the case where 
         //  the projection is on the segment
         if (t <= 0)
            closestPoint = lineSegmentPoint1;
         else if (t >= 1)
            closestPoint = lineSegmentPoint2;
         else 
            closestPoint = lineSegmentPoint1 + (lineSegment * t);
         return this.DistanceSquared(closestPoint);
      }


      public double DotProduct(MyVector otherVector)
      {
         return this._x * otherVector._x + this._y * otherVector._y;
      }

      public static MyVector operator +(MyVector leftVector, MyVector rightVector)
      {
         return new MyVector(leftVector._x + rightVector._x, leftVector._y + rightVector._y);
      }

      public static MyVector operator -(MyVector leftVector, MyVector rightVector)
      {
         return new MyVector(leftVector._x - rightVector._x, leftVector._y - rightVector._y);
      }

      public static MyVector operator *(MyVector aVector, double aScalar)
      {
         return new MyVector(aVector._x * aScalar, aVector._y * aScalar);
      }

      // Added using ReSharper due to CodeAnalysis nagging

      public bool Equals(MyVector other)
      {
         return _x.Equals(other._x) && _y.Equals(other._y);
      }

      public override bool Equals(object obj)
      {
         if (ReferenceEquals(null, obj)) return false;
         return obj is MyVector && Equals((MyVector) obj);
      }

      public override int GetHashCode()
      {
         unchecked
         {
            return (_x.GetHashCode()*397) ^ _y.GetHashCode();
         }
      }

      public static bool operator ==(MyVector left, MyVector right)
      {
         return left.Equals(right);
      }

      public static bool operator !=(MyVector left, MyVector right)
      {
         return !left.Equals(right);
      }
   }

这是一个小测试程序。

   public static class JustTesting
   {
      public static void Main()
      {
         Stopwatch stopwatch = new Stopwatch();
         stopwatch.Start();

         for (int i = 0; i < 10000000; i++)
         {
            TestIt(1, 0, 0, 0, 1, 1, 0.70710678118654757);
            TestIt(5, 4, 0, 0, 20, 10, 1.3416407864998738);
            TestIt(30, 15, 0, 0, 20, 10, 11.180339887498949);
            TestIt(-30, 15, 0, 0, 20, 10, 33.541019662496844);
            TestIt(5, 1, 0, 0, 10, 0, 1.0);
            TestIt(1, 5, 0, 0, 0, 10, 1.0);
         }

         stopwatch.Stop();
         TimeSpan timeSpan = stopwatch.Elapsed;
      }


      private static void TestIt(float aPointX, float aPointY, 
                                 float lineSegmentPoint1X, float lineSegmentPoint1Y, 
                                 float lineSegmentPoint2X, float lineSegmentPoint2Y, 
                                 double expectedAnswer)
      {
         // Katz
         double d1 = DistanceFromPointToLineSegment(new MyVector(aPointX, aPointY), 
                                              new MyVector(lineSegmentPoint1X, lineSegmentPoint1Y), 
                                              new MyVector(lineSegmentPoint2X, lineSegmentPoint2Y));
         Debug.Assert(d1 == expectedAnswer);

         /*
         // Katz using squared distance
         double d2 = DistanceFromPointToLineSegmentSquared(new MyVector(aPointX, aPointY), 
                                              new MyVector(lineSegmentPoint1X, lineSegmentPoint1Y), 
                                              new MyVector(lineSegmentPoint2X, lineSegmentPoint2Y));
         Debug.Assert(Math.Abs(d2 - expectedAnswer * expectedAnswer) < 1E-7f);
          */

         /*
         // Matti (optimized)
         double d3 = FloatVector.DistanceToLineSegment(new PointF(aPointX, aPointY), 
                                                new PointF(lineSegmentPoint1X, lineSegmentPoint1Y), 
                                                new PointF(lineSegmentPoint2X, lineSegmentPoint2Y));
         Debug.Assert(Math.Abs(d3 - expectedAnswer) < 1E-7f);
          */
      }

      private static double DistanceFromPointToLineSegment(MyVector aPoint, 
                                             MyVector lineSegmentPoint1, MyVector lineSegmentPoint2)
      {
         MyVector closestPoint;  // Not used
         return aPoint.DistanceToLineSegment(lineSegmentPoint1, lineSegmentPoint2, 
                                             out closestPoint);
      }

      private static double DistanceFromPointToLineSegmentSquared(MyVector aPoint, 
                                             MyVector lineSegmentPoint1, MyVector lineSegmentPoint2)
      {
         MyVector closestPoint;  // Not used
         return aPoint.DistanceToLineSegmentSquared(lineSegmentPoint1, lineSegmentPoint2, 
                                                    out closestPoint);
      }
   }

正如您所看到的，我尝试测量使用避免 Sqrt() 方法的版本和普通版本之间的差异。 我的测试表明您也许可以节省大约 2.5%，但我什至不确定 - 各种测试运行中的变化具有相同的数量级。 我还尝试测量 Matti 发布的版本（加上明显的优化），该版本似乎比基于 Katz/Grumdrig 代码的版本慢约 4%。

编辑：顺便说一句，我还尝试过测量一种使用叉积（和 Sqrt()）找到到无限直线（不是线段）的距离的方法，速度大约快 32%。

Answer 13

这是 devnullicus 的 C++ 版本转换为 C#。 对于我的实施，我需要知道交叉点并发现他的解决方案运行良好。

public static bool PointSegmentDistanceSquared(PointF point, PointF lineStart, PointF lineEnd, out double distance, out PointF intersectPoint)
{
    const double kMinSegmentLenSquared = 0.00000001; // adjust to suit.  If you use float, you'll probably want something like 0.000001f
    const double kEpsilon = 1.0E-14; // adjust to suit.  If you use floats, you'll probably want something like 1E-7f
    double dX = lineEnd.X - lineStart.X;
    double dY = lineEnd.Y - lineStart.Y;
    double dp1X = point.X - lineStart.X;
    double dp1Y = point.Y - lineStart.Y;
    double segLenSquared = (dX * dX) + (dY * dY);
    double t = 0.0;

    if (segLenSquared >= -kMinSegmentLenSquared && segLenSquared <= kMinSegmentLenSquared)
    {
        // segment is a point.
        intersectPoint = lineStart;
        t = 0.0;
        distance = ((dp1X * dp1X) + (dp1Y * dp1Y));
    }
    else
    {
        // Project a line from p to the segment [p1,p2].  By considering the line
        // extending the segment, parameterized as p1 + (t * (p2 - p1)),
        // we find projection of point p onto the line. 
        // It falls where t = [(p - p1) . (p2 - p1)] / |p2 - p1|^2
        t = ((dp1X * dX) + (dp1Y * dY)) / segLenSquared;
        if (t < kEpsilon)
        {
            // intersects at or to the "left" of first segment vertex (lineStart.X, lineStart.Y).  If t is approximately 0.0, then
            // intersection is at p1.  If t is less than that, then there is no intersection (i.e. p is not within
            // the 'bounds' of the segment)
            if (t > -kEpsilon)
            {
                // intersects at 1st segment vertex
                t = 0.0;
            }
            // set our 'intersection' point to p1.
            intersectPoint = lineStart;
            // Note: If you wanted the ACTUAL intersection point of where the projected lines would intersect if
            // we were doing PointLineDistanceSquared, then intersectPoint.X would be (lineStart.X + (t * dx)) and intersectPoint.Y would be (lineStart.Y + (t * dy)).
        }
        else if (t > (1.0 - kEpsilon))
        {
            // intersects at or to the "right" of second segment vertex (lineEnd.X, lineEnd.Y).  If t is approximately 1.0, then
            // intersection is at p2.  If t is greater than that, then there is no intersection (i.e. p is not within
            // the 'bounds' of the segment)
            if (t < (1.0 + kEpsilon))
            {
                // intersects at 2nd segment vertex
                t = 1.0;
            }
            // set our 'intersection' point to p2.
            intersectPoint = lineEnd;
            // Note: If you wanted the ACTUAL intersection point of where the projected lines would intersect if
            // we were doing PointLineDistanceSquared, then intersectPoint.X would be (lineStart.X + (t * dx)) and intersectPoint.Y would be (lineStart.Y + (t * dy)).
        }
        else
        {
            // The projection of the point to the point on the segment that is perpendicular succeeded and the point
            // is 'within' the bounds of the segment.  Set the intersection point as that projected point.
            intersectPoint = new PointF((float)(lineStart.X + (t * dX)), (float)(lineStart.Y + (t * dY)));
        }
        // return the squared distance from p to the intersection point.  Note that we return the squared distance
        // as an optimization because many times you just need to compare relative distances and the squared values
        // works fine for that.  If you want the ACTUAL distance, just take the square root of this value.
        double dpqX = point.X - intersectPoint.X;
        double dpqY = point.Y - intersectPoint.Y;

        distance = ((dpqX * dpqX) + (dpqY * dpqY));
    }

    return true;
}

Answer 14

2D 和 3D 解决方案

考虑更改基础，使线段变为 (0, 0, 0)-(d, 0, 0) 且点变为 >（u，v，0）。 最短距离出现在该平面中，由下式给出

    u ≤ 0 -> d(A, C)
0 ≤ u ≤ d -> |v|
d ≤ u     -> d(B, C)

（到端点之一或到支撑线的距离，具体取决于线的投影。等距轨迹由两个半圆和两条线段组成。 )

在上面的表达式中，d是线段AB的长度，u、v分别是AB/d（方向上的单位向量）的标量积和叉积（模） AB）和AC。 因此从向量上看，

AB.AC ≤ 0             -> |AC|
    0 ≤ AB.AC ≤ AB²   -> |ABxAC|/|AB|
          AB² ≤ AB.AC -> |BC|

Answer 15

C#版本

public static FP DistanceToLineSegment(FPVector3 a, FPVector3 b, FPVector3 point)
{
  var d = b - a;
  var s = d.SqrMagnitude;
  var ds = d / s;
  var lambda = FPVector3.Dot(point - a, ds);
  var p = FPMath.Clamp01(lambda) * d;
  return (a + p - point).Magnitude;
}

Answer 16

上述功能不适用于垂直线。 这是一个运行良好的函数！
与点 p1、p2 的直线。 检查点是p；

public float DistanceOfPointToLine2(PointF p1, PointF p2, PointF p)
{
  //          (y1-y2)x + (x2-x1)y + (x1y2-x2y1)
  //d(P,L) = --------------------------------
  //         sqrt( (x2-x1)pow2 + (y2-y1)pow2 )

  double ch = (p1.Y - p2.Y) * p.X + (p2.X - p1.X) * p.Y + (p1.X * p2.Y - p2.X * p1.Y);
  double del = Math.Sqrt(Math.Pow(p2.X - p1.X, 2) + Math.Pow(p2.Y - p1.Y, 2));
  double d = ch / del;
  return (float)d;
}

Answer 17

请参阅以下网站中的 Matlab GEOMETRY 工具箱：
http://people.sc.fsu.edu/~jburkardt/m_src /geometry/geometry.html

ctrl+f 输入“segment”即可查找线段相关函数。 您需要函数“segment_point_dist_2d.m”和“segment_point_dist_3d.m”。

GEOMETRY 代码有 C 版本和 C++ 版本、FORTRAN77 版本、FORTRAN90 版本和 MATLAB 版本。

Answer 18

基于 Joshua Javascript 的 AutoHotkeys 版本：

plDist(x, y, x1, y1, x2, y2) {
    A:= x - x1
    B:= y - y1
    C:= x2 - x1
    D:= y2 - y1

    dot:= A*C + B*D
    sqLen:= C*C + D*D
    param:= dot / sqLen

    if (param < 0 || ((x1 = x2) && (y1 = y2))) {
        xx:= x1
        yy:= y1
    } else if (param > 1) {
        xx:= x2
        yy:= y2
    } else {
        xx:= x1 + param*C
        yy:= y1 + param*D
    }

    dx:= x - xx
    dy:= y - yy

    return sqrt(dx*dx + dy*dy)
}

Answer 19

接受的答案不起作用
（例如，0,0 和 (-10,2,10,2) 之间的距离应为 2）。

这是有效的代码：

   def dist2line2(x,y,line):
     x1,y1,x2,y2=line
     vx = x1 - x
     vy = y1 - y
     ux = x2-x1
     uy = y2-y1
     length = ux * ux + uy * uy
     det = (-vx * ux) + (-vy * uy) #//if this is < 0 or > length then its outside the line segment
     if det < 0:
       return (x1 - x)**2 + (y1 - y)**2
     if det > length:
       return (x2 - x)**2 + (y2 - y)**2
     det = ux * vy - uy * vx
     return det**2 / length
   def dist2line(x,y,line): return math.sqrt(dist2line2(x,y,line))

Answer 20

这里没有看到 Java 实现，所以我将 Javascript 函数从已接受的答案翻译为 Java 代码：

static double sqr(double x) {
    return x * x;
}
static double dist2(DoublePoint v, DoublePoint w) {
    return sqr(v.x - w.x) + sqr(v.y - w.y);
}
static double distToSegmentSquared(DoublePoint p, DoublePoint v, DoublePoint w) {
    double l2 = dist2(v, w);
    if (l2 == 0) return dist2(p, v);
    double t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
    if (t < 0) return dist2(p, v);
    if (t > 1) return dist2(p, w);
    return dist2(p, new DoublePoint(
            v.x + t * (w.x - v.x),
            v.y + t * (w.y - v.y)
    ));
}
static double distToSegment(DoublePoint p, DoublePoint v, DoublePoint w) {
    return Math.sqrt(distToSegmentSquared(p, v, w));
}
static class DoublePoint {
    public double x;
    public double y;

    public DoublePoint(double x, double y) {
        this.x = x;
        this.y = y;
    }
}

Answer 21

Eli，您选择的代码不正确。 靠近线段所在线但远离线段一端的点将被错误地判断为靠近线段。 更新：提到的错误答案不再被接受。

这是一些正确的 C++ 代码。 它假定一个 2D 向量类 class vec2 {float x,y;}，本质上，具有加、减、缩放等运算符，以及距离和点积函数（即 x1 x2 + y1 y2）。

float minimum_distance(vec2 v, vec2 w, vec2 p) {
  // Return minimum distance between line segment vw and point p
  const float l2 = length_squared(v, w);  // i.e. |w-v|^2 -  avoid a sqrt
  if (l2 == 0.0) return distance(p, v);   // v == w case
  // Consider the line extending the segment, parameterized as v + t (w - v).
  // We find projection of point p onto the line. 
  // It falls where t = [(p-v) . (w-v)] / |w-v|^2
  // We clamp t from [0,1] to handle points outside the segment vw.
  const float t = max(0, min(1, dot(p - v, w - v) / l2));
  const vec2 projection = v + t * (w - v);  // Projection falls on the segment
  return distance(p, projection);
}

编辑：我需要一个 Javascript 实现，所以就在这里，没有依赖项（或注释，但它是上面的直接端口）。 点表示为具有 x 和 y 属性的对象。

function sqr(x) { return x * x }
function dist2(v, w) { return sqr(v.x - w.x) + sqr(v.y - w.y) }
function distToSegmentSquared(p, v, w) {
  var l2 = dist2(v, w);
  if (l2 == 0) return dist2(p, v);
  var t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
  t = Math.max(0, Math.min(1, t));
  return dist2(p, { x: v.x + t * (w.x - v.x),
                    y: v.y + t * (w.y - v.y) });
}
function distToSegment(p, v, w) { return Math.sqrt(distToSegmentSquared(p, v, w)); }

编辑2：我需要一个Java版本，但更重要的是，我需要3d而不是2d。

float dist_to_segment_squared(float px, float py, float pz, float lx1, float ly1, float lz1, float lx2, float ly2, float lz2) {
  float line_dist = dist_sq(lx1, ly1, lz1, lx2, ly2, lz2);
  if (line_dist == 0) return dist_sq(px, py, pz, lx1, ly1, lz1);
  float t = ((px - lx1) * (lx2 - lx1) + (py - ly1) * (ly2 - ly1) + (pz - lz1) * (lz2 - lz1)) / line_dist;
  t = constrain(t, 0, 1);
  return dist_sq(px, py, pz, lx1 + t * (lx2 - lx1), ly1 + t * (ly2 - ly1), lz1 + t * (lz2 - lz1));
}

这里，在函数参数中， 是有问题的点，线段的端点 和。 函数 dist_sq（假设存在）计算两点之间距离的平方。

Answer 22

这是最简单的完整 JavaScript 代码。

x, y 是您的目标点，x1, y1 到 x2, y2 是您的线段。

更新：修复评论中的 0 长度行问题。

function pDistance(x, y, x1, y1, x2, y2) {

  var A = x - x1;
  var B = y - y1;
  var C = x2 - x1;
  var D = y2 - y1;

  var dot = A * C + B * D;
  var len_sq = C * C + D * D;
  var param = -1;
  if (len_sq != 0) //in case of 0 length line
      param = dot / len_sq;

  var xx, yy;

  if (param < 0) {
    xx = x1;
    yy = y1;
  }
  else if (param > 1) {
    xx = x2;
    yy = y2;
  }
  else {
    xx = x1 + param * C;
    yy = y1 + param * D;
  }

  var dx = x - xx;
  var dy = y - yy;
  return Math.sqrt(dx * dx + dy * dy);
}

更新：Kotlin 版本

fun getDistance(x: Double, y: Double, x1: Double, y1: Double, x2: Double, y2: Double): Double {
    val a = x - x1
    val b = y - y1
    val c = x2 - x1
    val d = y2 - y1

    val lenSq = c * c + d * d
    val param = if (lenSq != .0) { //in case of 0 length line
        val dot = a * c + b * d
        dot / lenSq
    } else {
        -1.0
    }

    val (xx, yy) = when {
        param < 0 -> x1 to y1
        param > 1 -> x2 to y2
        else -> x1 + param * c to y1 + param * d
    }

    val dx = x - xx
    val dy = y - yy
    return hypot(dx, dy)
}

Answer 23

这是针对有限线段的实现，而不是像这里大多数其他函数那样的无限线（这就是我这样做的原因）。

Paul Bourke 理论的实现。

Python：

def dist(x1, y1, x2, y2, x3, y3): # x3,y3 is the point
    px = x2-x1
    py = y2-y1

    norm = px*px + py*py

    u =  ((x3 - x1) * px + (y3 - y1) * py) / float(norm)

    if u > 1:
        u = 1
    elif u < 0:
        u = 0

    x = x1 + u * px
    y = y1 + u * py

    dx = x - x3
    dy = y - y3

    # Note: If the actual distance does not matter,
    # if you only want to compare what this function
    # returns to other results of this function, you
    # can just return the squared distance instead
    # (i.e. remove the sqrt) to gain a little performance

    dist = (dx*dx + dy*dy)**.5

    return dist

AS3：

public static function segmentDistToPoint(segA:Point, segB:Point, p:Point):Number
{
    var p2:Point = new Point(segB.x - segA.x, segB.y - segA.y);
    var something:Number = p2.x*p2.x + p2.y*p2.y;
    var u:Number = ((p.x - segA.x) * p2.x + (p.y - segA.y) * p2.y) / something;

    if (u > 1)
        u = 1;
    else if (u < 0)
        u = 0;

    var x:Number = segA.x + u * p2.x;
    var y:Number = segA.y + u * p2.y;

    var dx:Number = x - p.x;
    var dy:Number = y - p.y;

    var dist:Number = Math.sqrt(dx*dx + dy*dy);

    return dist;
}

Java

private double shortestDistance(float x1,float y1,float x2,float y2,float x3,float y3)
    {
        float px=x2-x1;
        float py=y2-y1;
        float temp=(px*px)+(py*py);
        float u=((x3 - x1) * px + (y3 - y1) * py) / (temp);
        if(u>1){
            u=1;
        }
        else if(u<0){
            u=0;
        }
        float x = x1 + u * px;
        float y = y1 + u * py;

        float dx = x - x3;
        float dy = y - y3;
        double dist = Math.sqrt(dx*dx + dy*dy);
        return dist;

    }

Answer 24

在我自己的问题线程如何在 C、C# / .NET 2.0 或 Java 中计算所有情况下点和线段之间的最短 2D 距离？ 当我找到一个答案时，有人要求我在此处放置一个 C# 答案：所以这里是从 http://www.topcoder 修改的。 com/tc?d1=tutorials&d2=geometry1&module=Static ：

//Compute the dot product AB . BC
private double DotProduct(double[] pointA, double[] pointB, double[] pointC)
{
    double[] AB = new double[2];
    double[] BC = new double[2];
    AB[0] = pointB[0] - pointA[0];
    AB[1] = pointB[1] - pointA[1];
    BC[0] = pointC[0] - pointB[0];
    BC[1] = pointC[1] - pointB[1];
    double dot = AB[0] * BC[0] + AB[1] * BC[1];

    return dot;
}

//Compute the cross product AB x AC
private double CrossProduct(double[] pointA, double[] pointB, double[] pointC)
{
    double[] AB = new double[2];
    double[] AC = new double[2];
    AB[0] = pointB[0] - pointA[0];
    AB[1] = pointB[1] - pointA[1];
    AC[0] = pointC[0] - pointA[0];
    AC[1] = pointC[1] - pointA[1];
    double cross = AB[0] * AC[1] - AB[1] * AC[0];

    return cross;
}

//Compute the distance from A to B
double Distance(double[] pointA, double[] pointB)
{
    double d1 = pointA[0] - pointB[0];
    double d2 = pointA[1] - pointB[1];

    return Math.Sqrt(d1 * d1 + d2 * d2);
}

//Compute the distance from AB to C
//if isSegment is true, AB is a segment, not a line.
double LineToPointDistance2D(double[] pointA, double[] pointB, double[] pointC, 
    bool isSegment)
{
    double dist = CrossProduct(pointA, pointB, pointC) / Distance(pointA, pointB);
    if (isSegment)
    {
        double dot1 = DotProduct(pointA, pointB, pointC);
        if (dot1 > 0) 
            return Distance(pointB, pointC);

        double dot2 = DotProduct(pointB, pointA, pointC);
        if (dot2 > 0) 
            return Distance(pointA, pointC);
    }
    return Math.Abs(dist);
} 

我 @SO 不是为了回答而是提出问题，所以我希望我不会因为某些原因而获得百万票反对，而是构建评论家。 我只是想（并且受到鼓励）分享其他人的想法，因为该线程中的解决方案要么使用某种外来语言（Fortran、Mathematica），要么被某人标记为错误。 对我来说唯一有用的一个（由 Grumdrig 编写）是用 C++ 编写的，没有人标记它有错误。 但它缺少被调用的方法（点等）。

Answer 25

对于任何感兴趣的人，这里有一个 Joshua 的 Javascript 代码到 Objective-C 的简单转换：

- (double)distanceToPoint:(CGPoint)p fromLineSegmentBetween:(CGPoint)l1 and:(CGPoint)l2
{
    double A = p.x - l1.x;
    double B = p.y - l1.y;
    double C = l2.x - l1.x;
    double D = l2.y - l1.y;

    double dot = A * C + B * D;
    double len_sq = C * C + D * D;
    double param = dot / len_sq;

    double xx, yy;

    if (param < 0 || (l1.x == l2.x && l1.y == l2.y)) {
        xx = l1.x;
        yy = l1.y;
    }
    else if (param > 1) {
        xx = l2.x;
        yy = l2.y;
    }
    else {
        xx = l1.x + param * C;
        yy = l1.y + param * D;
    }

    double dx = p.x - xx;
    double dy = p.y - yy;

    return sqrtf(dx * dx + dy * dy);
}

我需要这个解决方案与 MKMapPoint 一起使用，所以我将分享它，以防其他人需要它。 只需进行一些细微的更改，这将返回以米为单位的距离：

- (double)distanceToPoint:(MKMapPoint)p fromLineSegmentBetween:(MKMapPoint)l1 and:(MKMapPoint)l2
{
    double A = p.x - l1.x;
    double B = p.y - l1.y;
    double C = l2.x - l1.x;
    double D = l2.y - l1.y;

    double dot = A * C + B * D;
    double len_sq = C * C + D * D;
    double param = dot / len_sq;

    double xx, yy;

    if (param < 0 || (l1.x == l2.x && l1.y == l2.y)) {
        xx = l1.x;
        yy = l1.y;
    }
    else if (param > 1) {
        xx = l2.x;
        yy = l2.y;
    }
    else {
        xx = l1.x + param * C;
        yy = l1.y + param * D;
    }

    return MKMetersBetweenMapPoints(p, MKMapPointMake(xx, yy));
}

Answer 26

在 F# 中，从点 c 到 a 和 b 之间的线段的距离由以下公式给出：

let pointToLineSegmentDistance (a: Vector, b: Vector) (c: Vector) =
  let d = b - a
  let s = d.Length
  let lambda = (c - a) * d / s
  let p = (lambda |> max 0.0 |> min s) * d / s
  (a + p - c).Length

向量 d 沿着线段从 a 指向 b。 d/s 与 ca 的点积给出了无限直线与点 c 之间最接近点的参数。 min 和 max 函数用于将此参数限制在 0..s 范围内，以便该点位于 a< 之间/code> 和 b。 最后，a+pc的长度就是c到线段上最近点的距离。

使用示例：

pointToLineSegmentDistance (Vector(0.0, 0.0), Vector(1.0, 0.0)) (Vector(-1.0, 1.0))

Answer 27

在 Mathematica 中，

它使用线段的参数化描述，并将点投影到线段定义的线上。 当参数在线段中从 0 变为 1 时，如果投影超出此范围，我们将计算到相应端点的距离，而不是与线段垂直的直线。

Clear["Global`*"];
 distance[{start_, end_}, pt_] := 
   Module[{param},
   param = ((pt - start).(end - start))/Norm[end - start]^2; (*parameter. the "."
                                                       here means vector product*)

   Which[
    param < 0, EuclideanDistance[start, pt],                 (*If outside bounds*)
    param > 1, EuclideanDistance[end, pt],
    True, EuclideanDistance[pt, start + param (end - start)] (*Normal distance*)
    ]
   ];  

绘制结果：

Plot3D[distance[{{0, 0}, {1, 0}}, {xp, yp}], {xp, -1, 2}, {yp, -1, 2}]

绘制那些比截止距离更近的点：

等值线图：

Answer 28

嘿，这是我昨天刚写的。 它位于 Actionscript 3.0 中，基本上是 Javascript，尽管您可能没有相同的 Point 类。

//st = start of line segment
//b = the line segment (as in: st + b = end of line segment)
//pt = point to test
//Returns distance from point to line segment.  
//Note: nearest point on the segment to the test point is right there if we ever need it
public static function linePointDist( st:Point, b:Point, pt:Point ):Number
{
    var nearestPt:Point; //closest point on seqment to pt

    var keyDot:Number = dot( b, pt.subtract( st ) ); //key dot product
    var bLenSq:Number = dot( b, b ); //Segment length squared

    if( keyDot <= 0 )  //pt is "behind" st, use st
    {
        nearestPt = st  
    }
    else if( keyDot >= bLenSq ) //pt is "past" end of segment, use end (notice we are saving twin sqrts here cuz)
    {
        nearestPt = st.add(b);
    }
    else //pt is inside segment, reuse keyDot and bLenSq to get percent of seqment to move in to find closest point
    {
        var keyDotToPctOfB:Number = keyDot/bLenSq; //REM dot product comes squared
        var partOfB:Point = new Point( b.x * keyDotToPctOfB, b.y * keyDotToPctOfB );
        nearestPt = st.add(partOfB);
    }

    var dist:Number = (pt.subtract(nearestPt)).length;

    return dist;
}

另外，这里对这个问题有一个非常完整且可读的讨论： notejot.com

Answer 29

使用反正切线的直线解决方案：

其想法是将A移动到(0, 0)并顺时针旋转三角形以使C位于X轴上，
当发生这种情况时，By 将是距离。

1. 角度 = Atan(Cy - Ay, Cx - Ax);
2. b 角度 = Atan(By - Ay, Bx - Ax);
3. AB 长度 = Sqrt( (Bx - Ax)^2 + (By - Ay)^2 )
4. By = Sin ( bAngle - aAngle) * ABLength

C#

public double Distance(Point a, Point b, Point c)
{
    // normalize points
    Point cn = new Point(c.X - a.X, c.Y - a.Y);
    Point bn = new Point(b.X - a.X, b.Y - a.Y);

    double angle = Math.Atan2(bn.Y, bn.X) - Math.Atan2(cn.Y, cn.X);
    double abLength = Math.Sqrt(bn.X*bn.X + bn.Y*bn.Y);

    return Math.Sin(angle)*abLength;
}

一行 C#（要转换为 SQL）

double distance = Math.Sin(Math.Atan2(b.Y - a.Y, b.X - a.X) - Math.Atan2(c.Y - a.Y, c.X - a.X)) * Math.Sqrt((b.X - a.X) * (b.X - a.X) + (b.Y - a.Y) * (b.Y - a.Y))

Answer 30

对于懒惰的人来说，这是我上面 @Grumdrig 解决方案的 Objective-C 端口：

CGFloat sqr(CGFloat x) { return x*x; }
CGFloat dist2(CGPoint v, CGPoint w) { return sqr(v.x - w.x) + sqr(v.y - w.y); }
CGFloat distanceToSegmentSquared(CGPoint p, CGPoint v, CGPoint w)
{
    CGFloat l2 = dist2(v, w);
    if (l2 == 0.0f) return dist2(p, v);

    CGFloat t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
    if (t < 0.0f) return dist2(p, v);
    if (t > 1.0f) return dist2(p, w);
    return dist2(p, CGPointMake(v.x + t * (w.x - v.x), v.y + t * (w.y - v.y)));
}
CGFloat distanceToSegment(CGPoint point, CGPoint segmentPointV, CGPoint segmentPointW)
{
    return sqrtf(distanceToSegmentSquared(point, segmentPointV, segmentPointW));
}